Well, this is about an innovative approach (innovative in the sense, I didn't see it anywhere before), to solve some **induction** problems using the **recurrence relations**.

We know that if you want to prove that \[\color{Purple}{(1+\sqrt{5})^n + (1-\sqrt{5})^n}\] is always an even integer, \(\forall n\in \mathbb{N}\), then what can we use ?

\(\mathbf{1.}\quad\)First is, you can use the binomial expansion formula, that is \(\displaystyle \color{Green}{ (1+x)^n = 1+\binom{n}{1} x + \binom{n}{2}x^2+...+\binom{n}{n}x^n = \sum_{k=0}^n \dbinom{n}{k} x^k }\)

Then, because in 2nd term, \(\sqrt{5}\) has a negative sign, in the sum of two brackets' expansion, the terms with an **odd power** of \(\sqrt{5}\) get cancelled and the terms with even power are added to each other twice (Once from \((1+\sqrt{5})^n\) and once from \((1-\sqrt{5})^n\) )

Hence it will be even integer.

\(\mathbf{2.} \quad\)Other way is induction. For \(n=1\), it's trivially true.

Assume for \(n=k\) and prove for \(n=k+1\).

\(\mathbf{3.} \quad\quad\) But the more interesting (seemingly interesting) one, which I thought of is as follows.

See that \(1+\sqrt{5}\) and \(1-\sqrt{5}\) are the roots of the quadratic equation \[x^2- 2x -4=0\]

Thus think of the recurrence relation which has it's characteristic equation \(x^2=2x+4\), and it is \[a_n=2a_{n-1} + 4a_{n-2}\] and with initial conditions, \(a_0=a_1=2\) (We chose these conditions so that we can later get general form as wanted)

Then we see that as the recurrence is starting with \(\textbf{integers}\), and recurrence has integer coefficients , every term will be \(\textbf{integers}\), and from the RHS, as it is \(2(a_{n-1}+2a_{n-2})\), it will always be even.

Now let's generalize the recurrence, and surely, because we designed the recurrence from quadratic whose roots were the \(1+\sqrt{5}\) and \(1-\sqrt{5}\)), and we chose the initial conditions accordingly, so the general term of this recurrence is **confirmly**

\(a_n=(1+\sqrt{5})^n+(1-\sqrt{5})^n\)

And hence, because each term of this recurrence is \(\color{Red}{\textbf{even integer}}\), we have proved that \((1+\sqrt{5})^n+(1-\sqrt{5})^n\) is an even integer \(\forall n\in \mathbb{N}\).

\(\color{Blue}{\textbf{Cool, isn't it?}}\)

And the advantage of this method is that this also proves that actually \((1+\sqrt{5})^n+(1-\sqrt{5})^n\) is always divisible by \(4\), which was not asked though, but an even more precise answer.

\(\color{Blue}{\textbf{Really, there's more than meets the eye in Maths....}}\)

If you never saw this approach before, and liked it then , like (brilliantwise :P) and reshare ;)

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## Comments

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TopNewestNice note, @Aditya Raut. :D

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Thanks, सर्व तुमचाच आशीर्वाद आहे .

\(\bullet \quad \quad\) Challenge to @Sharky Kesa and @Finn Hulse get without Google translate....

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I don't know Marathi. I used to, but that was long long time ago, when I was only 3. I can pronounce what you have written, however.

Sarv thumachaach aabhovadhee aahai.

Why did I get 3 down votes for that comment?

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for it.Log in to reply

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Arya

:)Log in to reply

I know Hindi because I am Indian. This name is an Internet pseudonym. :D

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@Sharky Kesa Sharky Sharky Sharky ! How many surprises am I going to receive from you before I go mad ?

m

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don'tinclude:I work for the CIA

I made my own business which owns every single other corporation on the planet

I am a girl

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@Sharky Kesa .... Seriously \(\color{Red}{\bigoplus \Box }\)

I really want to go to Jungle after reading the further surprises byLog in to reply

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I can read Hindi/Marathi, nextI am indian, and nowI am girl, I have company etc..... too much of surprises :PLog in to reply

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usedto be able to understand Marathi. I can pronounce Hindi characters. And I sad that the list was about what I am not.Log in to reply

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Wait we both speak and read Hindi though.

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@Aditya Raut, 'Tu te marathit kasa lihila?' Lol!

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Baraha's website . Hope that helps. @Ameya Salankar

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@Aditya Raut

वा! वा! मला तर हे माहितिच न्व्ह्त। Just Great!Log in to reply

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Excellent!

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For n=1. Answer is 2. Which is not divisible by 4..... I think it's divisible by 4 right from n=2...... :)

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Ok, that's good observation, but i didn't mean for them. See the approach I have told asks initial conditions which we got by n=0 and n=1 , and all else is talked about

furthersequence, but that point is correct though.... @Saikrishna JampuramLog in to reply

Can you explain what are you trying to tell.

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Before the last few lines from the explanation he is telling that term is always divisible by 4 ..... So if you substitute n=1 and check it..... It's not. :)

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@a d Nice note, :) :D ^^ @Aditya Raut u're the best!!

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Pleased to hear that, \(\color{Red}{\Huge{\textbf{Thank you very much!}}}\)

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You just had completed a first class Numerical Method course

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Didn't get you, sorry ?

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you can learn more methods just like this in Numerical Methods.Just google Numerical Methods

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Not Surprised, Because I have known it. Source: "Art of Problem Solving" by Authur

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Oh where? Please give the source, I didn't know it's already used somewhere. Thanks for sharing though ...

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